JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    Let \[f(x)=\frac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}-\frac{d-x}{\sqrt{{{b}^{2}}+{{(d-x)}^{2}}}},x\in R\]where a, b and d are non-zero real constants. Then [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) f' is not a continuous function of x

    B) f is neither increasing nor decreasing function of x

    C) f is an increasing function of x

    D) f is a decreasing function of x

    Correct Answer: C

    Solution :

    Here,\[f(x)=\frac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}-\frac{d-x}{\sqrt{{{b}^{2}}+{{(d-x)}^{2}}}}\] \[\Rightarrow \]\[f'(x)=\frac{{{a}^{2}}}{{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}+\frac{{{b}^{2}}}{{{({{b}^{2}}+{{(d-x)}^{2}})}^{3/2}}}>0\] \[\forall x\in R\] \[\therefore \] f(x) is an increasing function of x


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